A) Molecular weight of \[KMn{{O}_{4}}\]
B) \[\frac{1}{2}\]´ Molecular weight of \[KMn{{O}_{4}}\]
C) \[\frac{1}{3}\]´ Molecular weight of \[KMn{{O}_{4}}\]
D) \[\frac{1}{5}\]´ Molecular weight of \[KMn{{O}_{4}}\]
Correct Answer: D
Solution :
\[\frac{1}{5}\times \text{ molecular }\,\,\text{weigth of }KMn{{O}_{4}}\] as transfer of \[5{{e}^{-}}\] takes place when \[KMn{{O}_{4}}\] acts as oxidant in acidic medium. \[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O\]\[+5O\]
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